cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A327303 One of the two successive approximations up to 5^n for the 5-adic integer sqrt(-9). This is the 4 (mod 5) case (except for n = 0).

Original entry on oeis.org

0, 4, 4, 79, 79, 79, 3204, 18829, 331329, 1112579, 1112579, 20643829, 118300079, 850721954, 3292128204, 27706190704, 149776503204, 302364393829, 1065303846954, 8694698378204, 46841671034454, 332943965956329, 332943965956329, 5101315547987579, 28943173458143829
Offset: 0

Views

Author

Jianing Song, Sep 16 2019

Keywords

Comments

a(n) is the unique number k in [1, 5^n] and congruent to 4 mod 5 such that k^2 + 9 is divisible by 5^n.

Examples

			The unique number k in {4, 9, 14, 19, 24} such that k^2 + 9 is divisible by 25 is k = 4, so a(2) = 4.
The unique number k in {4, 29, 54, 79, 104} such that k^2 + 9 is divisible by 125 is k = 79, so a(3) = 46.
The unique number k in {79, 204, 329, 454, 579} such that k^2 + 9 is divisible by 625 is k = 79, so a(4) = 79.
		

Crossrefs

For the digits of sqrt(-9) see A327304 and A327305.
Approximations of 5-adic square roots:
A327302, this sequence (sqrt(-9));
A324027, A324028 (sqrt(-6));
A268922, A269590 (sqrt(-4));
A048898, A048899 (sqrt(-1));
A324023, A324024 (sqrt(6)).

Programs

  • Maple
    R:= [padic:-rootp(x^2+9,5,101)]:
    R:= op(select(t -> padic:-ratvaluep(t,1)=4, R)):
    seq(padic:-ratvaluep(R,n),n=0..100); # Robert Israel, Jan 16 2023
  • PARI
    a(n) = truncate(-sqrt(-9+O(5^n)))

Formula

a(1) = 4; for n >= 2, a(n) is the unique number k in {a(n-1) + m*5^(n-1) : m = 0, 1, 2, 3, 4} such that k^2 + 9 is divisible by 5^n.
For n > 0, a(n) = 5^n - A327302(n).

A269593 a(n) = (A268922(n)^2 + 4)/5^n, n >= 0.

Original entry on oeis.org

4, 1, 5, 1, 109, 1460, 292, 53476, 124904, 993169, 5385572, 43930532, 239139524, 777233593, 789206948, 2256445369, 65340851012, 661111023620, 132222204724, 7745142596633, 10225443529556, 103321258570120, 20664251714024, 4562022446935993, 6246398287209928, 20888388201358465
Offset: 0

Views

Author

Wolfdieter Lang, Mar 02 2016

Keywords

Comments

a(n) is an integer because b(n) = A268922(n) satisfies b(n)^2 + 4 == 0 (mod 5^n), n>=0.
See A268922 for details and references.

Examples

			a(0) = (0 + 4)/1 = 4.
a(4) = (261^2 + 4)/5^4 = 109.
		

Crossrefs

Cf. A268922, A269590, A269594 (companion sequence).

Programs

  • PARI
    a(n) = ((truncate(sqrt(-4+O(5^(n)))))^2 + 4)/5^n; \\ Michel Marcus, Mar 07 2016

Formula

a(n) = (b(n)^2 + 1)/5^n, n>=0, with b(n) = A268922(n).

Extensions

Terms a(21) and beyond from Andrew Howroyd, Mar 02 2020

A327302 One of the two successive approximations up to 5^n for the 5-adic integer sqrt(-9). This is the 1 (mod 5) case (except for n = 0).

Original entry on oeis.org

0, 1, 21, 46, 546, 3046, 12421, 59296, 59296, 840546, 8653046, 28184296, 125840546, 369981171, 2811387421, 2811387421, 2811387421, 460575059296, 2749393418671, 10378787949921, 48525760606171, 143893192246796, 2051241825059296, 6819613407090546, 30661471317246796
Offset: 0

Views

Author

Jianing Song, Sep 16 2019

Keywords

Comments

a(n) is the unique number k in [1, 5^n] and congruent to 1 mod 5 such that k^2 + 9 is divisible by 5^n.

Examples

			The unique number k in {1, 6, 11, 16, 21} such that k^2 + 9 is divisible by 25 is k = 21, so a(2) = 21.
The unique number k in {21, 46, 71, 96, 121} such that k^2 + 9 is divisible by 125 is k = 46, so a(3) = 46.
The unique number k in {46, 171, 296, 421, 546} such that k^2 + 9 is divisible by 625 is k = 546, so a(4) = 546.
		

Crossrefs

For the digits of sqrt(-9) see A327304 and A327305.
Approximations of 5-adic square roots:
this sequence, A327303 (sqrt(-9));
A324027, A324028 (sqrt(-6));
A268922, A269590 (sqrt(-4));
A048898, A048899 (sqrt(-1));
A324023, A324024 (sqrt(6)).

Programs

  • PARI
    a(n) = truncate(sqrt(-9+O(5^n)))

Formula

a(1) = 1; for n >= 2, a(n) is the unique number k in {a(n-1) + m*5^(n-1) : m = 0, 1, 2, 3, 4} such that k^2 + 9 is divisible by 5^n.
For n > 0, a(n) = 5^n - A327303(n).
Previous Showing 11-13 of 13 results.